Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>

int n,x,po[200001],f[200001],ans;

bool check(int a,int b,int c) {return a<b&&b<c||b<c&&c<a||c<a&&a<b;}

int main()
{	
	scanf("%d",&n); ans=0;
	for (int i=1; i<=n; i++) scanf("%d",&x),po[x]=i;
	for (int i=n-1; i>=1; i--)
		if (!ans) {if (po[i+1]<po[i]) ans++,f[i]=i+1;}
		else if (check(po[i],po[i+1],po[f[i+1]])) f[i]=f[i+1]; else f[i]=i+1,ans++;
	printf("%d\n",ans);
	return 0;
}